Ch.02 Modeling of Vibratory Systems
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Transcript of Ch.02 Modeling of Vibratory Systems
2/2/2014
1
02. Modeling of Vibratory
System
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Vibrations 2.01 Modeling of Vibratory Systems
Chapter Objectives
β’ Compute the mass moment of inertia of rotational systems
β’ Determine the stiffness of various linear and nonlinear elastic
components in translation and torsion and the equivalent
stiffness when many individual linear components are
combined
β’ Determine the stiffness of fluid, gas, and pendulum elements
β’ Determine the potential energy of stiffness elements
β’ Determine the damping for systems that have different sources
of dissipation: viscosity, dry friction, fluid, and material
β’ Construct models of vibratory systems
Vibrations 2.02 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§1.Introduction
- Three elements that comprise a vibrating system
β’ Inertia elements: stores and releases kinetic energy
β’ Stiffness elements: stores and releases potential energy
β’ Dissipation elements: express energy loss in a system
- Components comprising a vibrating mechanical system
β’ Translational motion β’ Rotational motion
Mass, π (ππ) Mass moment of inertia, π½ (πππ2)
Stiffness, π (π/π) Stiffness, ππ‘ (ππ/πππ)
Damping, π (ππ /π) Damping, ππ‘ (πππ /πππ)
External force, πΉ (π) External moment, π (ππ)
Vibrations 2.03 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§2.Inertia Elements
- Translational motion π
- Rotational motion
Slender bar π½πΊ =1
12ππΏ2
Circular disk π½πΊ =1
2ππ 2
Sphere π½πΊ =2
5ππ 2
Circular cylinder π½π₯ = π½π¦ =1
12π(3π 2 + β2)
π½π§ =1
2ππ 2
π½π = π½πΊ + ππ2, π: distance from the center of gravity to point πΊ
Vibrations 2.04 Modeling of Vibratory Systems
- Parallel-axes theorem
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§2.Inertia Elements
- For a mass π translating with a
velocity of magnitude π₯ in the πβπ
plane under the driving force πΉ
β’ The equation governing the motion of the mass π
πΉ π =π
ππ‘(π π₯ π)
when π and π are independent of time
πΉ = π π₯ (2.2)
β’ The kinetic energy, π, of mass π
π =1
2π π₯ π β π₯ π =
1
2π π₯2
Vibrations 2.05 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§2.Inertia Elements
- For a rigid body undergoing only
rotation in the plane with an angular
speed π
β’ The equation governing the rotation of the mass of inertia
π = π½ π (2.6)
π : the moment acting about the center of mass πΊ or a
fixed point π along the direction normal to the plane of
motion
π½ : the associated mass moment of inertia
β’ The kinetic energy of the system
π =1
2π½ π2
Vibrations 2.06 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
2
Β§2.Inertia Elements
- Ex.2.1 Determination of mass moments of inertia
Illustrate how the mass moments of inertia of several different
rigid body distributions are determined
Solution
β’ Uniform Disk
The mass moment of inertia about the point π ,
which is located a distance π from point πΊ
π½π = π½πΊ + ππ 2 =1
2ππ 2 + ππ 2 =
3
2ππ 2
β’ Uniform Bar
The mass moment of inertia about the point π
π½π = π½πΊ + ππΏ
2
2
=1
12ππΏ2 +
1
4ππΏ2 =
1
3ππΏ2
Vibrations 2.07 Modeling of Vibratory Systems
π½πΊ =1
2ππ 2
π½πΊ =1
12ππΏ2
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§2.Inertia Elements
- Ex.2.2 Slider mechanism: system with varying inertia property
A slider of mass ππ slides along a uniform bar of
mass ππ with a pivot at point π. Another bar,
which is pivoted at point πβ², has a portion of length
π that has a mass ππ and another portion of
length π that has a mass ππ. Determine the rotary
inertia π½π of this system and show its dependence
on the angular displacement coordinate π
From geometry, π, ππ, ππ can be described in terms of π
π2 π = π2 + π2 β 2πππππ π
ππ2 π = (π/2)2+π2 β πππππ π
ππ2 π = (π/2)2+π2 β πππππ (π β π)
ππ : the distance from the midpoint of bar of mass ππ to π
ππ : the distance from the midpoint of bar of mass ππ to π
Vibrations 2.08 Modeling of Vibratory Systems
Solution
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§2.Inertia Elements
π2 π = π2 + π2 β 2πππππ π
ππ2 π = (π/2)2+π2 β πππππ π
ππ2 π = (π/2)2+π2 β πππππ (π β π)
The rotary inertia π½π of this system
π½π = π½ππ+ π½ππ
(π) + π½ππ(π) + π½ππ
(π)
where
π½ππ=
1
3πππ
2
π½ππ (π) = ππ π
2(π)
π½πππ = ππ
π2
12+ ππππ
2 = ππ
π2
3+ π2 β πππππ π
π½πππ = ππ
π2
12+ ππππ
2 = ππ
π2
3+ π2 β πππππ (π β π)
Vibrations 2.09 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
1.Introduction
- Stiffness elements are manufactured from different materials
and they have many different shapes
- Application
β’ to minimize vibration transmission from machinery to the
supporting structure
β’ to isolate a building from earthquakes
β’ to absorb energy from systems subjected to impacts
Vibrations 2.10 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Some representative types of stiffness elements that are
commercially available along with their typical application
Vibrations 2.11 Modeling of Vibratory Systems
Building or highway
base isolation for
lateral motion using
cylindrical rubber
bearings
Wire rope isolators to isolate vertical motions of machinery
Steel cable
springs
used in a
chimney
tuned
mass
damper to
suppress
lateral
motions
Air springs used in
suspension systems to
isolate vertical motions
Typical steel
coil springs
for isolation
of vertical
motions
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- The stiffness elements store and release the potential energy
of a system
- Consider a spring under acting force of magnitude πΉ is
directed along the direction of the unit vector π
πΉπ = βπΉ π
πΉπ tries to restore the stiffness element to its undeformed
configuration, it is referred to as a restoring force
Vibrations 2.12 Modeling of Vibratory Systems
Stiffness element with a force acting on it Free-body diagram
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
3
Β§3.Stiffness Elements
- As the stiffness element is deformed, energy is stored in this
element, and as the stiffness element is undeformed, energy
is released
- The potential energy π is defined as the work done to take
the stiffness element from the deformed position to the
undeformed position; that is, the work needed to undeform
the element to its original shape
π π₯ = ππππππππ πππ ππ‘πππ
ππππ‘πππ ππ πππππππππ πππ ππ‘πππ
πΉπ ππ₯
=
π₯
0
πΉπ ππ₯ =
π₯
0
βπΉ π β ππ₯ π =
π₯
0
πΉππ₯
Vibrations 2.13 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
2.Linear Springs
- Translation Spring
β’ Deformation
πΉ π₯ = ππ₯ (2.9)
πΉ : the applied force
π : the spring constant
π₯ : the spring deflection
β’ The potential energy π stored in the spring
π π₯ =
0
π₯
πΉ π₯ ππ₯ =
0
π₯
ππ₯ππ₯ = π
0
π₯
π₯ππ₯ =1
2ππ₯2
Vibrations 2.14 Modeling of Vibratory Systems
(2.10)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Torsion Spring
β’ Deformation
π π = ππ‘π (2.11)
π : the applied moment
ππ‘ : the spring constant
π : the spring deflection
β’ The potential energy π stored in the spring
π π =
0
π
π π ππ =
0
π
ππ‘πππ =1
2ππ‘π
2
Vibrations 2.15 Modeling of Vibratory Systems
(2.12)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Combinations of Linear Springs
β’ Parallel Springs
Translation springs
Total force
πΉ π₯ =πΉ1 π₯ +πΉ1 π₯ =π1π₯+π2π₯= π1 +π2 π₯ =πππ₯
Equivalent spring
ππ = π1 + π2
Torsion springs
Total moment
π π = π1 π + π2 π
= ππ‘1π + ππ‘2
π = ππ‘1+ ππ‘2
π = ππ‘ππ
Equivalent spring
ππ‘π= ππ‘1
+ ππ‘2
Vibrations 2.16 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ Series Springs
Translation springs
π₯ = π₯1 + π₯2 =πΉ
π1+
πΉ
π2=
1
π1+
1
π2πΉ =
πΉ
ππ
Equivalent spring
ππ =1
π1+
1
π2
β1
=π1π2
π1 + π2
Torsion springs
π = π1 + π2 =π
ππ‘1
+π
ππ‘2
=1
ππ‘1
+1
ππ‘2
π =π
ππ‘π
Equivalent spring
ππ‘π=
1
ππ‘1
+1
ππ‘2
β1
=ππ‘1
ππ‘2
ππ‘1+ ππ‘2
Vibrations 2.17 Modeling of Vibratory Systems
Displacement
Displacement
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Spring Constants for Some Common Elastic Elements
1. Axially loaded rod or cable
π =π΄πΈ
πΏ
π΄ : cross-sectional area, π2
πΈ : Youngβs modulus of elasticity, π/π2
πΏ : length of the rod, π
2. Axially loaded tapered rod
π =ππΈπ1π2
4πΏ
πΈ : Youngβs modulus of elasticity, π/π2
π1 : rod diameter, π
π2 : rod diameter, π
πΏ : length of the rod, π
Vibrations 2.18 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
4
Β§3.Stiffness Elements
3. Hollow circular rod in torsion
ππ‘ =πΊπΌ
πΏ
πΊ : shear modulus of elasticity, π/π2
πΌ : the torsion constant (polar moment of inertia), π4
For the concentric circular tubes,
πΏ : length of the rod, π
ππ : outside rod diameter, π
ππ : inside rod diameter, π
4. Cantilever beam
π =3πΈπΌ
π3, 0 < π β€ πΏ
πΈ : Youngβs modulus of elasticity, π/π2
πΌ : the area moment of inertia about the bending axis, π4
π : position of applied force, π
πΏ : length of the beam, π
Vibrations 2.19 Modeling of Vibratory Systems
πΌ =π(ππ
4 β ππ4)
32
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
5. Pinned-pinned beam (Hinged, simply supported)
π =3πΈπΌ(π + π)
π2π2
πΈ : Youngβs modulus of elasticity, π/π2
πΌ : the area moment of inertia about the bending axis, π4
π,π: position of applied force, π
6. Clamped-clamped beam (Fixed-fixed beam)
π =3πΈπΌ(π + π)3
π3π3
πΈ : Youngβs modulus of elasticity, π/π2
πΌ : the area moment of inertia about the bending axis, π4
π,π: position of applied force, π
Vibrations 2.20 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
7. Two circular rods in torsion
ππ‘π= ππ‘1
+ ππ‘2, ππ‘π
=πΊππΌππΏπ
πΊπ : modulus of elasticity, π/π2
πΌπ : the torsion constant (polar moment of inertia), π4
πΏπ : position of applied force, π
8. Two circular rods in torsion
ππ‘π=
1
ππ‘1
+1
ππ‘2
β1
, ππ‘π=
πΊππΌππΏπ
πΊπ : modulus of elasticity, π/π2
πΌπ : the torsion constant (polar moment of inertia), π4
πΏπ : position of applied force, π
Vibrations 2.21 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
9. Coil springs
π =πΊπ4
8ππ·3
πΊ : modulus of elasticity, π/π2
π : wire diameter, π
π : number of active coil
π· : mean coil diameter, π
10. Clamped rectangular plate, constant thickness, force at center
π =πΈβ3
12πΌπ2(1 β π2)
πΈ : Youngβs modulus of elasticity, π/π2
β : thickness of plate, π
πΌ : coefficient
π : width of the plate, π
π : poison ratio
Vibrations 2.22 Modeling of Vibratory Systems
π/π πΌ1.0 0.005601.2 0.006471.4 0.006911.6 0.007121.8 0.007202.0 0.00722
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Force-displacement relationships may also be used to
determine parameters such as π that characterize a stiffness
element
Vibrations 2.23 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.3 Equivalent stiffness of a beam-spring combination
Consider a cantilever beam that has a spring attached at its free end
β’ The force is applied to the free end of the spring
ππ =1
πππππ+
1
π1
β1
πππππ =3πΈπΌ
πΏ3
β’ The force is applied simultaneously to the free end of the
cantilever beam
ππ = πππππ + π1
Vibrations 2.24 Modeling of Vibratory Systems
βΉ parrallel
βΉ series
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
5
Β§3.Stiffness Elements
- Ex.2.4 Equivalent stiffness of a cantilever beam with a
transverse end load
Acantilever beam:πΈ = 72Γ109π/π2, π = 750ππ, ππ = 110ππ,
ππ = 120ππ. Determine the equivalent stiffness of this beam
Solution
The area moment of inertia πΌ about the bending axis
πΌ =π
32ππ
4 β ππ4
=π
32120 Γ 10β3 4 β 110 Γ 10β3 4
= 5.98 Γ 10β6π4
The equivalent stiffness of the cantilever beam
π =3πΈπΌ
πΏ3 =3 Γ 72 Γ 109 Γ 5.98 Γ 10β6
750 Γ 10β3 3 = 3.06 Γ 106π/π
Vibrations 2.25 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.5 Equivalent stiffness of a beam with a fixed end and a
translating support at the other end
Consider a uniform beam of length πΏ with flexural rigidity πΈπΌ. When the
beam is subjected to a transverse loading πΉat the translating support end, determine
the equivalent stiffness of this beam
Solution
By observation
βΊ
Vibrations 2.26 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
To this end, we use Case 6 of Table 2.3 and set π = π = πΏ and
obtain
ππππ₯ππ = 3πΈπΌ π + π 3
π3π3
π=π=πΏ
=3πΈπΌ πΏ + πΏ 3
πΏ3πΏ3 =24πΈπΌ
πΏ3
Recognizing that the equivalent stiffness of a fixed-fixed beam
of length 2πΏ loaded at its middle is equal to the total equivalent
stiffness of a parallel spring combination of two end loaded
beams, we obtain that
ππ =1
2ππππ₯ππ =
1
2
24πΈπΌ
πΏ3 =12πΈπΌ
πΏ3
Vibrations 2.27 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.6 Equivalent stiffness of a micro-electromechanical
system (MEMS) fixed-fixed flexure
A micro-electromechanical sensor
system (MEMS) consisting of four
flexures. Each of these flexures is
fixed at one end and connected to a
mass at the other end. The length of each flexure isπΏ =100ππ, the thickness of each flexure is β = 2ππ, and the
width of each flexure is π = 2ππ. A transverse loading acts on
the mass along the π-direction, which is normal to the π β πplane. Each flexure is fabricated from a poly-silicon material,
which has a Youngβs modulus of elasticity πΈ = 150πΊππ
Determine the equivalent stiffness of the system
Vibrations 2.28 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
Solution
Each of the four flexures can be treated as a beam that is fixed
at one end and free to translate only
at the other end, similar to the
system in Ex.2.5
The equivalent stiffness of each
flexure is given by
πππππ₯π’ππ =12πΈπΌ
πΏ3 , πΌ =πβ3
12The equivalent stiffness of the system
ππ = 4 Γ πππππ₯π’ππ = 4 Γ12πΈ Γ
πβ3
12πΏ3 = 4
πΈπβ3
πΏ3
= 4150 Γ 109 Γ 2 Γ 10β6 Γ 2 Γ 10β6
100 Γ 10β6 = 9.6π/π
Vibrations 2.29 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.7 Equivalent stiffness of springs in parallel: removal of a restriction
Determine of the equivalence spring constant for
the parallel springs subjected to unequal forces
Solution
From similar triangles
π₯ = π₯2 +π
π + ππ₯1 β π₯2 =
π
π + ππ₯1 +
π
π + ππ₯2
Consider the bar
πΉ = πΉ1 + πΉ2
ππΉ2 = ππΉ1
Therefore
π₯1 =πΉ1
π1=
ππΉ
π1 π + π, π₯2 =
πΉ2
π2=
ππΉ
π2 π + π
Vibrations 2.30 Modeling of Vibratory Systems
βΉ πΉ1 =ππΉ
π + π, πΉ2 =
ππΉ
π + π
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
6
Β§3.Stiffness Elements
βΉ π₯ =π
π + ππ₯1 +
π
π + ππ₯2
=π
π + π
ππΉ
π1 π + π+
π
π + π
ππΉ
π2 π + π
=πΉ
π + π 2
π1π2 + π2π
2
π1π2
For the equivalent system
πΉ = πππ₯
βΉ ππ =πΉ
π₯The equivalence spring constant for the parallel
springs subjected to unequal forces
ππ =π1π2 π + π 2
π1π2 + π2π
2
Vibrations 2.31 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
3.Nonlinear Springs
- Nonlinear stiffness elements appear in many applications,
including leaf springs in vehicle suspensions and uniaxial
micro-electromechanical devices in the presence of
electrostatic actuation
- The spring force πΉ(π₯)
πΉ π₯ = ππ₯ππππππ π πππππ πππππππ‘
+ πΌππ₯3
πππππππππ π πππππ πππππππ‘
(2.23)
πΌ : the stiffness coefficient of the nonlinear term
πΌ > 0 hardening spring πΌ < 0 softening spring
π : the linear spring constant
- The potential energy π
π π₯ = 0
π₯
πΉ π₯ ππ₯ =1
2ππ₯2 +
1
2πΌππ₯4
Vibrations 2.32 Modeling of Vibratory Systems
(2.24)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
πΉ π₯ = ππ₯ + πΌππ₯3 (2.23)
Β§3.Stiffness Elements
- For a nonlinear stiffness element described by Eq. (2.23), the
graph is no longer a straight line. The slope of this graph at a
location π₯ = π₯π is given by
ππΉ
ππ₯π₯=π₯π
= π + 3πΌππ₯2
π₯=π₯π
= π + 3πΌππ₯π2
βΉ in the vicinity of displacements in a neighborhood of π₯ =π₯π, the cubic nonlinear stiffness element may be replaced by
a linear stiffness element with a stiffness constant (2.25)
- The constant of proportionality πΌπ for the nonlinear cubic
spring is determined experimentally
Vibrations 2.33 Modeling of Vibratory Systems
(2.25)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Experimentally obtained data used to determine the
nonlinear spring constant πΌπ
Vibrations 2.34 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.8 Nonlinear stiffness due to geometry
a. Nonlinear stiffness due to geometry
β’ The initial tension force
π0 = πΉπ πΎ=0
= ππΏ0
β’ The force in the spring
πΉπ π₯ = ππΏ0 + π πΏ2 + π₯2 β πΏ
β’ The force in the π₯-direction is obtained
πΉπ₯ π₯ = πΉπ π πππΎ =πΉπ π₯
πΏ2 + π₯2=
π₯ππΏ0
πΏ2 + π₯2+
ππ₯ πΏ2 + π₯2 β πΏ
πΏ2 + π₯2
βΉ the spring force opposing the motion is a nonlinear function
of the displacement π₯ . Hence, this vibratory model of the
system will have nonlinear stiffness
Vibrations 2.35 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Binomial expansion 1 + π₯ π = 1 + ππ₯ +1
2π(π β 1)π₯2 + β―
Β§3.Stiffness Elements
Cubic Springs and Linear Springs
Assume that |π₯/πΏ| βͺ 1, using binomial expansion
1 +π₯
πΏ
2
= 1 +π₯
πΏ
2 1/2
= 1 +1
2
π₯
πΏ
2
+1
8
π₯
πΏ
4
+ β―
βΉ πΉπ₯ π₯ = ππΏ0
π₯/πΏ
1 + π₯/πΏ 2+
ππ₯ 1 + π₯/πΏ 2 β 1
1 + π₯/πΏ 2
= ππΏ0
π₯/πΏ
1 +12
π₯πΏ
2 +ππ₯ 1 +
12
π₯πΏ
2β 1
1 +12
π₯πΏ
2
= ππΏ0
π₯
πΏ+
π
2πΏ
π₯
πΏ
3
Vibrations 2.36 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
7
Β§3.Stiffness Elements
πΉπ₯ π₯ = ππΏ0
π₯
πΏ+
π
2πΏ
π₯
πΏ
3
When the nonlinear term is negligible
πΉπ₯ π₯ = ππΏ0
π₯
πΏ= π0
π₯
πΏand the spring constant is proportional to the initial tension in
the spring
b. Nonlinear spring composed of a set of linear springs
Another example of a nonlinear spring is one that is piecewise
linear as shown in figure
Vibrations 2.37 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
4.Other Forms of Potential Energy Elements
Consider other stiffness elements in which there is a
mechanism for storing and releasing potential energy. The
source of the restoring force is a fluid element or a gravitational
loading
- Fluid Element
β’ The magnitude of the total force of the
displaced fluid acting on the rest of the fluid
πΉπ π₯ = 2πππ΄0π₯
π : the mass density of the fluid, ππ/π3
π : gravitational constant, π/π 2
π΄0 : the manometer cross-sectional area, π2
π₯ : the fluid displacement, π
Vibrations 2.38 Modeling of Vibratory Systems
Manometer
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ The equivalent spring constant of this fluid system
ππ =ππΉπ
ππ₯= 2πππ΄0
β’ The potential energy
π π₯ =1
2πππ₯
2 = πππ΄0π₯2
Alternatively, the potential energy can also be obtained
directly from the work done
π π₯ = 0
π₯
πΉπ π₯ ππ₯
= 2πππ΄0 0
π₯
π₯ππ₯
= πππ΄0π₯2
Vibrations 2.39 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Compressed Gas
β’ When the piston moves by an amount π₯ along
the axis of the piston, π0 decreases to a
volume ππ
ππ π₯ = π0 β π΄0π₯ = π΄0πΏ0 1 β π₯/πΏ0
βΉ ππ π₯ = π0 1 β π₯/πΏ0
π΄0: the piston cross-sectional area, π2
β’ The equation of state for the gas
ππππ = π0π0
π = π0 = ππππ π‘ βΉ π = π0ππβπ
π : gas pressure, π/π2 π : gas volume, π3
π : the ratio of specific heats of the gas, when compressed
- slowly, the compression is isothermal, π = 1
- rapidly, the compression is adiabatic, π = ππ/ππ£ = 1.4
Vibrations 2.40 Modeling of Vibratory Systems
Gas compression
with a piston
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ The magnitude of the force on the piston
πΉ π₯ = π΄0π
= π΄0π0ππβπ
= π΄0π0π0βπ 1 β π₯/πΏ0
βπ
βΉ πΉ π₯ = π΄0π0 1 β π₯/πΏ0βπ (2.33)
The Eq. (2.33) describes a nonlinear force
versus displacement relationship
At the vicinity of π₯ = π₯π, the stiffness of an equivalent linear
stiffness element
ππ = ππΉ
ππ₯π₯=π₯π
=ππ΄0π0
πΏ01 β π₯π/πΏ0
βπβ1
For π₯π/πΏ0 βͺ 1,
Vibrations 2.41 Modeling of Vibratory Systems
ππ =ππ΄0π0
πΏ0
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ For arbitrary π₯/πΏ0, the potential energy
π π₯ = 0
π₯
πΉ π₯ ππ₯
= π΄0π0 0
π₯
1 β π₯/πΏ0βπππ₯
βΉ π π₯ = βπ΄0π0πΏ0ππ 1 β π₯/πΏ0 π = 1
π΄0π0πΏ0
π β 11 β π₯/πΏ0
βπ β 1 π β 1
Vibrations 2.42 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
8
Β§3.Stiffness Elements
- Pendulum System
β’ Pendulum systems: bar with uniformly
distributed mass
At π, the vertical distance through which the
center of gravity of the bar moves up from the
reference position
π₯ =πΏ
2β
πΏ
2πππ π =
πΏ
21 β πππ π
The increase in the potential energy
π π₯ = 0
π₯
πΉ π₯ ππ₯ = 0
π₯
ππππ₯ = πππ₯
or in π
π π =1
2πππΏ(1 β πππ π)
Vibrations 2.43 Modeling of Vibratory Systems
Pendulum systems: bar
with uniformly distributed
mass
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Taylor series approximation πππ π = 1 βπ2
2+ β―
Β§3.Stiffness Elements
π π =1
2πππΏ 1 β πππ π
β1
2
πππΏ
2π2
=1
2πππ
2
where the equivalent spring constant
ππ =πππΏ
2
Vibrations 2.44 Modeling of Vibratory Systems
Pendulum systems: bar
with uniformly distributed
mass
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ Pendulum systems: mass on a weightless rod
The increase in the potential energy
π π β1
2π1ππΏπ2 =
1
2πππ
2
where the equivalent spring constant
ππ = π1ππΏ
If the weightless bar is replaced by one that
has a uniformly distributed mass π, then the
total potential energy of the bar and the mass
π π β1
4πππΏπ2 +
1
2π1ππΏπ2 =
1
2
π
2+ π1 ππΏπ2 =
1
2πππ
2
where the equivalent spring constant
ππ =π
2+ π1 ππΏ
Vibrations 2.45 Modeling of Vibratory Systems
Pendulum systems: mass
on a weightless rod
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
β’ Pendulum systems: inverted mass on a
weightless rod
The decrease in the potential energy
π π β β1
2π1ππΏπ2
Vibrations 2.46 Modeling of Vibratory Systems
Pendulum systems:
inverted mass on a
weightless rod
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§3.Stiffness Elements
- Ex.2.9 Equivalent stiffness due to gravity loading
For βsmallβ rotations about the upright position ΞΈ = 0,
the potential energy
π π =1
2π1πππ2 β
1
2π2πππ2
=1
2π1π β π2π ππ2
There is a gain or loss in potential energy depending
on whether π1π > π2π or vice versa
β’ When the bar has a uniformly distributed mass π
π1 = ππΏ1/πΏ, π2 = ππΏ2/πΏ, πΏ = πΏ1 + πΏ2, π = πΏ1/2, π = πΏ2/2
π π =πΏ12 β πΏ2
2
4(πΏ1 + πΏ2)πππ2 =
1
2
πΏ1 + πΏ2
2πππ2 =
1
2πππ
2
where the equivalent stiffness ππ = πΏ1 + πΏ2 ππ/2
Vibrations 2.47 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
- Damping elements are assumed to have neither inertia nor the
means to store or release potential energy
- The mechanical motion imparted to these elements is
converted to heat or sound and, hence, they are called non-
conservative or dissipative because this energy is not
recoverable by the mechanical system
- There are four common types of damping mechanisms used to
model vibratory systems
β’ Viscous damping
β’ Coulomb or dry friction damping
β’ Material or solid or hysteretic damping
β’ Fluid damping
In all these cases, the damping force is expressed as a
function of velocity
Vibrations 2.48 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
9
Β§4.Dissipation Elements
1.Viscous damping
- When a viscous fluid flows through a slot or around a piston in
a cylinder, the damping force generated is proportional to the
relative velocity between the two boundaries confining the fluid
- A common representation of a viscous damper is a cylinder
with a piston head
- Depending on the damper construction and the velocity
range, the magnitude of the damper force πΉ( π₯) is a nonlinear
function of velocity or can be approximated as a linear
function of velocity
Vibrations 2.49 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
- In the linear case, the damper force is expressed as
πΉ π₯ = π π₯ (2.46)
π : the damping coefficient, π/(π/π )
Viscous damping of the form given by Eq. (2.46) is also
called slow-fluid damping
- In the case of a nonlinear viscous damper described by a
function πΉ π₯ , the equivalent linear viscous damping around
an operating speed π₯ = π₯π is determined as follows
ππ = ππΉ π₯
π π₯ π₯= π₯π
- Linear viscous damping elements can be combined in the
same way that linear springs are, except that the forces are
proportional to velocity instead of displacement
Vibrations 2.50 Modeling of Vibratory Systems
(2.47)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
β’ Energy Dissipation
The energy dissipated by a linear viscous damper
πΈπ = πΉππ₯ = πΉ π₯ππ‘ = π π₯2ππ‘ = π π₯2ππ‘
β’ Parallel-Plate Damper
An example of a viscous damper is shown in the figure
The shear force acting on the
bottom plate
πΉ π₯ =π π₯
βπ΄ =
ππ΄
β π₯
The damping coefficient π for the parallel-plate construction
π =ππ΄
β
Vibrations 2.51 Modeling of Vibratory Systems
(2.48)
(2.49)
(2.50)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
π =ππ΄
β(2.50)
Β§4.Dissipation Elements
- Ex.2.10 Design of a parallel-plate damper
A parallel-plate damper with a top plate of dimensions100ππΓ 100ππ is to be pulled across an oil layer of thickness
0.2ππ, which is confined between the moving plate and a
fixed plate. We are given that this oil is SAE30 oil, which has a
viscosity of 345ππππ (345 Γ 103ππ /π2)
Determine the viscous damping coefficient of this system
Solution
To this end, using Eq. (2.50) and substitute the given values
into this expression and find that
π =ππ΄
β=
345Γ10β3 100Γ10β3 Γ100Γ10β3
0.2 Γ 10β3 = 17.25ππ /π
Vibrations 2.52 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
- Ex.2.11 Equivalent damping coefficient and
equivalent stiffness of a vibratory system
Consider the vibratory system in which the motion of mass πis restrained by a set of linear springs and linear viscous
dampers. Determine ππ and ππ
Solution
The equivalence system
ππ = π1 +π2π3
π2 + π3, ππ = π1 + π2
Vibrations 2.53 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ππ = ππΉ π₯
π π₯ π₯= π₯π
(2.47)
Β§4.Dissipation Elements
- Ex.2.12 Equivalent linear damping coefficientof a nonlinear damper
It has been experimentally determined that the damper force-
velocity relationship is given by the function
πΉ π₯ = 4ππ /π π₯ + (0.3ππ 3/π) π₯3
Determine the equivalent linear damping coefficient around an
operating speed of 3π/π
Solution
Using the Eq. (2.47)
ππ = ππΉ π₯
π π₯ π₯=3π/π
= 4 + 0.9 π₯2
π₯=3π/π = 4 + 0.9 Γ 32
βΉ ππ = 12.1ππ /π
Vibrations 2.54 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
10
Β§4.Dissipation Elements
2.Other Forms of Dissipation
- Coulomb Damping or Dry Friction
This type of damping is due to the force caused by friction
between two solid surfaces
The friction force acting on the system
πΉ π₯ = πππ ππ( π₯) (2.51)
π : the kinetic coefficient of friction
π : the force compressing the surfaces, π
π ππ: the signum function, π ππ π₯ = +1 π₯ > 0β1 π₯ < 0
0 π₯ = 0
Vibrations 2.55 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
If the normal force is due to the system weight, then π = ππ
πΉ π₯ = ππππ ππ( π₯) (2.52)
The energy dissipated in this case
(2.53)
- Fluid Damping (Velocity-Squared Damping)
This type of damping is associated with a system whose mass is
vibrating in a fluid medium. The magnitude of the damping force
πΉ π₯ = ππ π₯2π ππ π₯ = ππ| π₯| π₯ (2.54)
ππ : friction coefficient, ππ = πΆππ΄/2
πΆ : drag coefficient
π : the mass density of the fluid
π΄ : the projected area of the mass in a direction normal to π₯
Fluid damping of (2.54) is often referred to as fast-fluid damping
Vibrations 2.56 Modeling of Vibratory Systems
πΈπ = πΉππ₯ = πΉ π₯ππ‘ = πππ π ππ π₯ π₯ππ‘
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§4.Dissipation Elements
The energy dissipated
πΈπ = πΉππ₯ = πΉ π₯ππ‘ = ππ π ππ( π₯) π₯3ππ‘
- Structural or Solid or Hysteretic Damping
This type of damping describes the losses in materials due to
internal friction. The damping force is a function of
displacement and velocity and is of the form
πΉ = πππ½βπ ππ π₯ |π₯| (2.57)
π½β : an empirically determined constant
The energy dissipated
πΈπ = πΉππ₯ = πΉ π₯ππ‘ = πππ½β π ππ( π₯)|π₯|ππ‘
Vibrations 2.57 Modeling of Vibratory Systems
(2.54)
(2.58)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§5.Model Construction
1.Introduction
- In this section, four examples are provided to illustrate how
the previously described inertia, stiffness, and damping
elements are used to construct system models
- Modeling is an art, and often experience serves as a guide in
model construction
- In this section, the examples are drawn from different areas,
and are presented in a progressive order proceeding from
the use of discrete inertia, stiffness, and damping elements in
a model to distributed elements, and finally, to a combination
of distributed and discrete elements
- As discussed in the subsequent chapters, the mass,
stiffness, and damping of a system appear as parameters in
the governing equations of the system
Vibrations 2.58 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§5.Model Construction
2.A Micro-electromechanical System
Micro-electromechanical accelerometer and a vibratory model
of this sensor
Vibrations 2.59 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§5.Model Construction
3.The Human Body
Human body and a vibratory model
Vibrations 2.60 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
11
Β§5.Model Construction
4.A Ski
Cross-country ski, which is a physical system with distributed
stiffness and inertia properties, and its vibratory model
Vibrations 2.61 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§5.Model Construction
5.Cutting Process
Work-piece-tool turning system and vibratory model of this
system
Vibrations 2.62 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Β§6.Design for Vibration
Vibrations 2.63 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Exercises
Vibrations 2.66 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien